Question

Tungsten trioxide (WO3) has a rich yellow color and is often used as a pigment in ceramics and paints. To test a ceramic vase for its WO2 content, a 10.20 g sample of the vase was ground and reduced with Pb(Hg) to convert any WO2 to W3+. The resulting W3+ was transferred to 500.0 mL of 1.00 M HCl. A 100.00 mL aliquot of the HCl solution required 12.30 mL of 0.09562 M potassium permanganate (KMnO4) to reach the purple endpoint. A blank required 0.09 mL Balance the redox reaction Mn2+Wo2 redox reaction: MnO^ W3+ Determine the percent WO3 in the ceramic sample percent WO3:

Answer 1

The balanced redox reaction is : 5 W³⁺ + 3 MnO₄^- + 4 H⁺ $\rightarrow$ 5 WO₂²⁺ + 3 Mn²⁺ + 2 H₂O

percent WO₃ : 22.11 %

Explanation

volume KMnO₄ required by aliquot = 12.30 mL

volume KMnO₄ required by blank = 0.09 mL

actual volume of KMnO₄ required = (volume KMnO₄ required by aliquot) - (volume KMnO₄ required by blank)

actual volume of KMnO₄ required = (12.30 mL) - (0.09 mL)

actual volume of KMnO₄ required = 12.21 mL

moles KMnO₄ required = (concentration KMnO₄) * (volume KMnO₄ required in Liter)

moles KMnO₄ required = (0.09562 M) * (12.21 x 10^-3 L)

moles KMnO₄ required = 1.1675 x 10^-3 mol

moles W³⁺ in aliquot = (moles KMnO₄ required) * (5 moles W³⁺ / 3 moles MnO₄^-)

moles W³⁺ in aliquot = (1.1675 x 10^-3 mol) * (5/3)

moles W³⁺ in aliquot = 1.946 x 10^-3 mol

Total moles W³⁺ = (moles W³⁺ in aliquot) * (total volume HCl / volume of aliquot)

Total moles W³⁺ = (1.946 x 10^-3 mol) * (500 mL / 100 mL)

Total moles W³⁺ = (1.946 x 10^-3 mol) * 5

Total moles W³⁺ = 9.73 x 10^-3 mol

moles WO₃ = Total moles W³⁺

moles WO₃ = 9.73 x 10^-3 mol

mass WO₃ = (moles WO₃) * (molar mass WO₃)

mass WO₃ = (9.73 x 10^-3 mol) * (231.8382 g/mol)

mass WO₃ = 2.2556 g

mass percent WO₃ = (mass WO₃ / mass ssample) * 100

mass percent WO₃ = (2.2556 g / 10.20 g) * 100

mass percent WO₃ = 22.11 %